114. Flatten Binary Tree to Linked List

Given the root of a binary tree, flatten the tree into a "linked list":

• The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
• The "linked list" should be in the same order as a pre-order traversal of the binary tree.

 

Example 1:

Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [0]
Output: [0]

 

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -100 <= Node.val <= 100

 

Follow up: Can you flatten the tree in-place (with O(1) extra space)?

Rust Solution

struct Solution;
use rustgym_util::*;

trait Postorder {
    fn postorder(self, prev: &mut TreeLink);
}

impl Postorder for TreeLink {
    fn postorder(self, prev: &mut TreeLink) {
        if let Some(node) = self {
            let left = node.borrow_mut().left.take();
            let right = node.borrow_mut().right.take();
            right.postorder(prev);
            left.postorder(prev);
            node.borrow_mut().right = prev.take();
            *prev = Some(node);
        }
    }
}

impl Solution {
    fn flatten(root: &mut TreeLink) {
        let mut prev: TreeLink = None;
        root.take().postorder(&mut prev);
        *root = prev;
    }
}

#[test]
fn test() {
    let mut root = tree!(1, tree!(2, tree!(3), tree!(4)), tree!(5, None, tree!(6)));
    let res = tree!(
        1,
        None,
        tree!(
            2,
            None,
            tree!(3, None, tree!(4, None, tree!(5, None, tree!(6))))
        )
    );
    Solution::flatten(&mut root);
    assert_eq!(root, res);
}